organic compounds containing nitrogen multiple choice ... - AWS

1

ORGANIC COMPOUNDS CONTAINING NITROGEN

MULTIPLE CHOICE QUESTIONS

1.

NNBF4 +

A

In the above process product A is

(a) Fluorobenzene (b) Benzene

(c) 1,4-difluorobenzene (d) 1,3-difluorobenzene

Sol. (a)

This is Baltz Schiemann reaction.

2. Treatment of NH3 with excess of ethyl chloride gives:

(a) diethylamine (b) ethane

(c) tetraethylammonium chloride (d) methylamine

Sol. (c)

4C2H5–Cl + NH3 C2H5–N–C2H5 Cl–

+

C2H5

C2H5

2

3. The product of the given reaction,

CH3 – C – C – CH3 HNO2

CH3 CH3

OH NH2 Product

is

(a)

CH3 – C – C – CH3

CH3 CH3

OH OH

(b) O

CH3– C – C – CH3

CH3

CH3

(c)

CH3 – C – CH – CH3

CH3

OH CH3

(d) CH3 – C = C – CH3

CH3 CH3

Sol. (b)

CH3 – C – C – CH3 HNO2

CH3 CH3

OH NH2

CH3 – C – C – CH3

CH3 CH3

OH

CH3 – C – C – CH3

CH3

OH

CH3

CH3 – C – C – CH3

CH3 O

CH3

–H+

4. Which among the following amines will give carbylamine

reaction?

(a) CH3–CH2–NH2 (b) CH3–NH–CH3

(c) (C6H5)3N (d) CH3–CH2–NH–OH

Sol. (a)

Carbyl amine test is responded by only 1°

aliphatic/aromatic amines.

5. Following 1° amine has chiral carbon as indicated CH3

H–C*–NH2

C6H5

.

This on reaction with (NaNO2 + HCl) forms:

3

(a) 1° alcohol with retention of configuration

(b) 2° alcohol with inverted configuration

(c) racemic mixture of 2° alcohol

(d) racemic mixture of 1° alcohol

Sol. (c)

Proceeds with SN1 stable benzylic carbocation.

6. The major product of the following reaction is

NH2

CH(OCH3)2

NaNO2

aq. H2SO4, Product

(a)

OH

CH(OCH3)2

(b)

NH2

CH(OH)2

(c)

OH

CHO

(d)

CHO

Sol. (c)

NH2

NaNO2

N2

CH(OCH3)2 aq. H2SO4

+

CH(OH2)

H2O,

OH

CHO

4

7. Which of the following product will be obtained in the

following reaction?

CH2NH2 HNO2

0–5°C ?

(a) CH2OH

(b) CH–OH

CH3

(c)

OH

(d) CH3CH2CH2CH2CH2OH

Sol. (c)

CH2–NH2 CH2–NH2–NO ¨ +

+

CH2–N2 +

CH2

OH

H2O

+ NO

8. Which of the following amine will show carbylamine

reaction?

(a)

NH2

CH3

(b)

CH3–N–H

(c)

N–CH3

CH3

(d)

NHCH3

CH3

Sol. (a)

Carbylamine reaction is given by primary amines only.

5

9. Reduction of nitrobenzene in the presence of Zn/NH4Cl

gives

(a) hydrazobenzene (b) aniline

(c) azobenzene (d) N-phenyl hydroxyl amine

Sol. (d)

+ 4[H]

NO2

Zn/NH4Cl

NHOH

+ H2O

Nitrobenzene N–Phenylhydroylamine

10. Which among the following amines will give carbylamine

reaction?

(a) CH3–CH2–NH2 (b) CH3–NH–CH3

(c) (C6H5)3N (d) CH3–CH2–NH–OH

Sol. (a)

Carbylamine test is responded by only 1° aliphatic/aromatic

amines.

11. Aniline reacts with excess of phosgene and KOH to form

(a) OH

(b)

C–Cl

O

(c)

NH–C–Cl

O

(d) N=C=O

Sol. (d)

C6H5NH2 + COCl2 KOH2 C6H5–N=C=O + 2KCl + 2H2O.

6

12. Which of the indicated ‘H’ is most acidic?

O

Cl O

H H

N

H

H H

N H

(A) (B)

(C)

(D)

(a) (a) (b) (b)

(c) (c) (d) (d)

Sol. (a)

13. The major product formed in the elimination reaction of

CH3

N

CH3

CH3

OH is

(a)

CH3 CH3

CH3 N

(b)

CH3 CH3

CH2 N

(c)

CH3 CH3

N

(d)

CH3 CH3

CH3 N

Sol. (b)

The least substituted alkene is formed as the major product

by Hoffmann’s elimination.

14. When CH3CH2CHCl2 is treated with NaNH2, the product

formed is

(a) CH3–CH = CH2 (b) CH3–C CH

(c) CH3CH2CH NH2

NH2 (d) CH3CH2CH

Cl

NH2

Sol. (b)

7

CH3–CH2–CH–Cl

Cl

CH3–CCH NaNH2

15. Which of the following is not the metamer of CH3CH2NCH2CH2CH2CH3 ?

CH3

(a) CH3NCH2CH2CH2CH2CH3

CH3

(b) CH3NCH2CHCH2CH2CH3

H CH3

(c) CH3NCH2CH2CH

CH3

CH3

CH3 (d) CH3CH2CH2NCH2CH2CH3

CH3

Sol. (b)

Compound (b) is not the metamer of the given compound

as it has only two alkyl groups attached to the Natom

whereas the given compound has three alkyl groups

attached to the nitrogen.

16. The base with lowest pKa value is

(a) NCCH2NH2 (b) Et3N

(c) NH3 (d) HOCH2CH2NH2

Sol. (c)

Low value of pKa means stronger acid or weaker base.

Thus as basicity decreases pKa value decreases. Due to

presence of strong electron withdrawing (I effect),

NCCH2NH2 is the weakest base as it has the lowest

electron density on Natom. Hence, NCCH2CH2 has the

lowest pKa value.

8

17. Which one of the following amines is the weakest base?

(a) NH2 (b) CH2NH2

(c)

NH2

CH3

(d) NHCH3

Sol. (c)

NH2

CH3

is the weakest base as its conjugate acid is the least

stable because of steric repulsion experienced by extra

proton and existing CH3 group at ortho position.

18. Which one of the following compounds is most basic in

aqueous medium?

(a) CH3CNH2

O

(b) CH3CNH2

S

(c) CH3CNH2

NH

(d) CH3CNH2

CH2

Sol. (c)

In aqueous medium

CH3CNH2 + H+

NH ..

Conjugate base

CH3CNH2 ..

NH2 +

..

(I)

CH3C=NH2

NH2

+

..

Conjugate acid

(Highly stable due to identical resonating structures)

(II)

9

19 CONH2

D

A mixture

and

CONH2

15

was subjected to Hofmann's bromamide degradation.

No body could ever

NH2

D

and

NH2

15

The products obtained are always a mixture of

report the formation of even the trace of the mixture of

NH2

and

NH2

D

15

(a) Hofmann’s rearrangement is intermolecular.

(b) Hofmann’s rearrangement is intramolecular.

(c) Hofmann’s rearrangement occurs very rapidly.

(d)Hofmann’s rearrangement occurs very slowly.

Sol. (b)

Hofmann’s rearrangement (i.e. shifting of Ph from C to

electron deficient N) is intramolecular, implies that Ph

never becomes free from the molecule, had it been so a

phenyl group from one molecule could have become

attached to the nitrogen of another giving crossproducts.

Hence, (b) is the correct answer.

20 Organic compound 5 2PCl H O/H

3 2 2 3rearrangement(A) B CH CH NH CH COOH

The compound (a) is

(a) synethylmethyl ketoxime

(b) antiethyl methyl ketoxime

(c) Nmethyl propionamide

10

(d) Nethyl acetamide

Sol. (a)

Hydrolysis products of (b) suggest that the Nsubstituted

amide is 3 2 3CH CONHCH CH . Hence,

3 2CH CH must be anti to OH

group of oxime (a). That is, the lighter alkyl 3CH is on the

side of OH group. Hence, compound (a) is synethyl

methyl ketoxime.

Hence, (a) is the correct answer.

21.

3 2 CH NH Product O

CH3

O

product is

(a) O

N

CH3

(b) O

NH

CH3

CH3

OH

(c) CH3

NH

CH3

O

OH

(d) None of the above

11

Sol.

O

CH3

O

CH3NH - HH

O

O

CH3

CH3NH

H

CH3

NH

O

OH

CH3

Hence, (c) is the correct answer.

22. Identify the end product (c).

3

2

CH I moist

excess Ag OA B C

N

H

(a)

(b

)

(c) N

+

CH3 CH3

(d

)

Sol. 32CH I

N

H

N+

CH3 CH3

2 2Ag O / H O

N+

CH3 CH3

H

I-

N

CH3 CH3

2 2Ag O / H O

N+

CH3 CH3CH3

I-

3CH I

N+

CH3 CH3CH3

H

penta 1,3 diene

(More stable)

OH-

Hence, (a) is the correct answer.

12

23. A compound (a) when reacted with PCl5 and then with NH3

gave (b), (b) when treated with Br2 and KOH produced (c).

(c) on treatment with NaNO2 and HCl at 0°C and on

boiling with water forms o-cresol. Compound A is

(a) o-toluic acid (b) o-chlorotoluene

(c) o-bromo toluene (d) m-toluic acid

Sol.

CH3

boil

OH2

OH

CH3

N2Cl

Diazatisation

CH3

NH2

Cl

KOHBr2

CH3

CONH2

(i) PCl5

(ii) NH3

CH3

COOH

(A)

(C) (B)

Hence, (d) is the correct answer.

24. C3H9N cannot represent

(a) 1o amine (b) 2° amine

(c) 3° amine (d) quaternary salt

Sol. (d)

General formula for any amine is CnH2n+3N; also note that

for primary amine, it is CnH2n+1NH2, for secondary amine it

is CnH2n+2NH and for tertiary amine it is CnH2n+3N. The

formula for quaternary salt is R4N+X–.

13

25 The correct order of basicities of the following compounds

is

CH3

NH2

NH

CH3

NH2

NH

CH3

CH3

CH3

NH2

O

1 2 3 4

(a) 2 1 3 4 (b) 1 3 2 4

(c) 3 1 2 4 (d) 1 2 3 4

Sol. (b)

The resonance structures of

2

3CH C NH|

NH

contains –ve charge

on one of the three nitrogen atoms, hence it is most basic.

3 2

O||

CH C NH will be least basic due to the delocalization of

electrons on nitrogen atom through resonance.

INTEGER TYPE QUESTIONS

26. How many primary amines are possible with formula

C4H11N?

Sol. 4

27. Certain nitrogenous compound with molecular mass (180)

show an increase in its molecular mass to 348 after

treatment with acetyl chloride. The number of possible –

NH2 groups in the molecule is:

sol. (4)

14

The chemical reaction involved is as follows:

3

|| ||

2 3 3

H

CH CO

O O

NH Cl C CH NH C CH

Net increase in mol. Mass on acylation of one –NH2 group

= Mol. Mass of CH3CO group – At. Mass of H

= 43 – 1 = 42

Actual increase in mol. Mass on acylation

= 348 – 180 = 168

no. CH3CO group added = 168/42 = 4

Hence the compound has 4 – NH2 groups.

28. A nitrogenous compound with molecular mass 180 shows

an increase in molecular mass to 348 after treatment with

acetyl chloride. The number of possible 2NH group in the

molecule is.

SOL. (4)

3

H

CH CO

3

CH

O

CNH

Increase in molecular mass on acylation of one 2NH group

is 42. Increase in Molecular mass = 348-180 = 168

No. of 2NH group 1684

42

29. Among R–CH2–NO2, Ph–NO2, Ph–CH2–NO2, R2CH–NO2

and R3C–NO2, how many compounds with give blue colour

when treated with HNO2 followed by NaOH

SOL. (1)

3CH

O

2NH C+ Cl

15

only R2CH–NO2 will give this reaction

30. Of the following amines how many can be separated by

Hoffmann's mustard oil reaction?

NH2, N

CH3

H

,

NH2

, N

CH3

Ph

, ,

CH3

NH2Ph Ph CH2 NH2

N

H

Ph

Sol. (4)

Primary amine gives positive test for Hoffmann's mustard

oil reaction.